Easy
Maximum Subarray With Equal Products — C++
Full explanation · Time precompute: O(r * log(logr)) runtime: O(n * log(logr)) · Space O(r * log(logr))
// Time: precompute: O(r * log(logr)), r = MAX_NUM
// runtime: O(n * log(logr))
// Space: O(r * log(logr))
// number theory, hash table
vector<vector<int>> prime_divisors(int n) {
const auto& linear_sieve_of_eratosthenes = [](int n) { // Time: O(n), Space: O(n)
vector<int> spf(n + 1, -1);
vector<int> primes;
for (int i = 2; i <= n; ++i) {
if (spf[i] == -1) {
spf[i] = i;
primes.emplace_back(i);
}
for (const auto& p : primes) {
if (i * p > n || p > spf[i]) {
break;
}
spf[i * p] = p;
}
}
return primes;
};
vector<vector<int>> result(n + 1);
for (const auto& p : linear_sieve_of_eratosthenes(n)) { // Time: O(nlog(logn))
for (int i = p; i <= n; i += p) {
result[i].emplace_back(p);
}
}
return result;
}
const int MAX_NUM = 10;
const auto& PRIME_DIVISORS = prime_divisors(MAX_NUM);
class Solution {
public:
int maxLength(vector<int>& nums) {
int result = 2;
unordered_map<int, int> lookup;
for (int right = 0, left = 0; right < size(nums); ++right) {
for (const auto& p : PRIME_DIVISORS[nums[right]]) {
left = max(left, lookup[p]);
lookup[p] = right + 1;
}
result = max(result, right - left + 1);
}
return result;
}
};